Rotation 4 Question 4
5. A thin ring of mass $2 kg$ and radius $0.5 m$ is rolling without slipping on a horizontal plane with velocity $1 m / s$. A small ball of mass $0.1 kg$, moving with velocity $20 m / s$ in the opposite direction, hits the ring at a height of $0.75 m$ and goes vertically up with velocity $10 m / s$. Immediately after the collision,
(2011)
(a) the ring has pure rotation about its stationary $CM$
(b) the ring comes to a complete stop
(c) friction between the ring and the ground is to the left
(d) there is no friction between the ring and the ground
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Answer:
Correct Answer: 5. $(a, c)$
Solution:
- The data is incomplete. Let us assume that friction from ground on ring is not impulsive during impact.
From linear momentum conservation in horizontal direction, we have
$$ \begin{array}{rlr} (-2 \times 1)+ & (0.1 \times 20) & \quad-ve \\ & =(0.1 \times 0)+(2 \times v) & +ve \end{array} $$
Here, $v$ is the velocity of $CM$ of ring after impact. Solving the above equation, we have $v=0$
Thus, CM becomes stationary.
$\therefore$ Correct option is (a).
Linear impulse during impact
(i) In horizontal direction
$$ J _1=\Delta p=0.1 \times 20=2 N-s $$
(ii) In vertical direction $J _2=\Delta p=0.1 \times 10=1 N$-s Writing the equation (about $CM$ )
Angular impulse $=$ Change in angular momentum
We have,
$$ 1 \times \frac{\sqrt{3}}{2} \times \frac{1}{2}-2 \times 0.5 \times \frac{1}{2}=2 \times(0.5)^{2} \quad \omega-\frac{1}{0.5} $$
Solving this equation $\omega$ comes out to be positive or $\omega$ anti-clockwise. So just after collision rightwards slipping is taking place.
Hence, friction is leftwards.
Therefore, option (c) is also correct.
